MHT CET · Physics · Current Electricity
Resistances in the left gap and right gap of a meter bridge are \(10 \Omega\) and \(30 \Omega\) respectively. If the resistances in the two gaps are interchanged, the balance point will shift to right by
- A 30 cm .
- B 40 cm .
- C 50 cm .
- D 60 cm .
Answer & Solution
Correct Answer
(C) 50 cm .
Step-by-step Solution
Detailed explanation
In first case
\(\begin{aligned}
& \mathrm{R}=10 \Omega \quad \mathrm{~S}=30 \Omega \\
& \frac{\mathrm{R}}{\mathrm{~S}}=\frac{l}{100-l} \\
& \frac{10}{30}=\frac{l}{100-l} \\
& l_1=25 \mathrm{~cm}
\end{aligned}\)
In second case
\(\begin{aligned}
& \mathrm{R}=30 \Omega ; \quad \mathrm{S}=10 \Omega \\
& \frac{\mathrm{R}}{\mathrm{~S}}=\frac{l^{\prime}}{100-l^{\prime}} \\
& \frac{30}{10}=\frac{l^{\prime}}{100-l^{\prime}} \\
& l^{\prime}=75 \mathrm{~cm}
\end{aligned}\)
\(\therefore \quad\) The balance point will shift by \(75 \mathrm{~cm}-25 \mathrm{~cm}\) is 50 cm
\(\begin{aligned}
& \mathrm{R}=10 \Omega \quad \mathrm{~S}=30 \Omega \\
& \frac{\mathrm{R}}{\mathrm{~S}}=\frac{l}{100-l} \\
& \frac{10}{30}=\frac{l}{100-l} \\
& l_1=25 \mathrm{~cm}
\end{aligned}\)
In second case
\(\begin{aligned}
& \mathrm{R}=30 \Omega ; \quad \mathrm{S}=10 \Omega \\
& \frac{\mathrm{R}}{\mathrm{~S}}=\frac{l^{\prime}}{100-l^{\prime}} \\
& \frac{30}{10}=\frac{l^{\prime}}{100-l^{\prime}} \\
& l^{\prime}=75 \mathrm{~cm}
\end{aligned}\)
\(\therefore \quad\) The balance point will shift by \(75 \mathrm{~cm}-25 \mathrm{~cm}\) is 50 cm
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